#### Answer

A cubic function has one point of inflection.
The x-coordinate of the inflection point is $~~\frac{x_1+x_2+x_3}{3}$

#### Work Step by Step

$y = ax^3+bx^2+cx+d$
$y' = 3ax^2+2bx+c$
$y'' = 6ax+2b$
We can find the value of $x$ such that $y'' = 0$:
$y'' = 6ax+2b = 0$
$x = -\frac{b}{3a}$
$y$ has one point of inflection and the x-coordinate is $x = -\frac{b}{3a}$
Suppose $y = 0$
$y = ax^3+bx^2+cx+d = 0$
$x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a} = 0$
Suppose $x_1, x_2,$ and $x_3$ are x-intercepts.
$(x-x_1)(x-x_2)(x-x_3) = 0$
$(x^2-x_1x-x_2x+x_1x_2)(x-x_3) = 0$
$x^3-x_1x^2-x_2x^2+x_1x_2x-x_3x^2+x_1x_3x+x_2x_3x-x_1x_2x_3 = 0$
$x^3-x_1x^2-x_2x^2-x_3x^2+x_1x_2x+x_1x_3x+x_2x_3x-x_1x_2x_3 = 0$
$x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3 = 0$
By comparing the two equations, we can see that $\frac{b}{a} = -(x_1+x_2+x_3)$
The x-coordinate of the inflection point is:
$x = -\frac{b}{3a}$
$x = -\frac{1}{3}(\frac{b}{a})$
$x = -\frac{1}{3}[-(x_1+x_2+x_3)]$
$x = \frac{x_1+x_2+x_3}{3}$
The x-coordinate of the inflection point is $~~\frac{x_1+x_2+x_3}{3}$